Question

Brantley tosses a ball out of a window that is 75 feet in the air. Its initial velocity is 12 feet per second. The path of the ball is represented by h=-16t^2+12t+75. How long does it take for the ball to hit the ground to the nearest hundredth?

Answer 1

Answer:

$t = 2.57$

Step-by-step explanation:

Given

$h=-16t^2+12t+75$

Required

Determine the time to hit the ground

The ground is at point 0.

So, to solve this; we simply set $h =0$, then calculate the value of t

$0=-16t^2+12t+75$

Multiply through by -1

$0=16t^2-12t-75$

Reorder

$16t^2-12t-75 = 0$

Solve using quadratic formula:

$t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}$

Where

$a = 16; b = -12; c = -75$

$t = \frac{-(-12)\±\sqrt{(-12)^2 - 4*16*(-75)}}{2*16}$

$t = \frac{12\±\sqrt{144 +4800}}{32}$

$t = \frac{12\±\sqrt{4944}}{32}$

$t = \frac{12\±70.31}{32}$

Split:

$t = \frac{12+70.31}{32}$ or $t = \frac{12-70.31}{32}$

$t = \frac{82.31}{32}$ or $t = \frac{-58.31}{32}$

But time (t) can't be negative;

So, we make use of only

$t = \frac{82.31}{32}$

$t = 2.57$

Hence, the time to hit the ground is 2.57 seconds